When a long capillary tube of radius 0.015 cm is dipped in a liquid, the liquid rises to a height of 15 cm within it. If the contact angle between the liquid and glass to close to `0^@`, the surface tension of the liquid, in milliNewton `m^(-1)`, is `[rho_("liquid") = 900 kgm^(-3), g = 10 ms^(-2)]` (give anwer is closet integer)

Explanation

To solve the problem, we will use the formula for the height of liquid rise in a capillary tube, which is given by:

\[ h = \frac{2T \cos \theta}{\rho g r} \]

Where:
- \( h \) = height of the liquid column (in meters)
- \( T \) = surface tension of the liquid (in N/m)
- \( \theta \) = contact angle (in degrees)
- \( \rho \) = density of the liquid (in kg/m³)
- \( g \) = acceleration due to gravity (in m/s²)
- \( r \) = radius of the capillary tube (in meters)

### Step 1: Convert the given values to appropriate units
- Radius \( r = 0.015 \, \text{cm} = 0.00015 \, \text{m} \)
- Height \( h = 15 \, \text{cm} = 0.15 \, \text{m} \)
- Density \( \rho = 900 \, \text{kg/m}^3 \)
- Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)
- Contact angle \( \theta \approx 0^\circ \) (thus \( \cos \theta = 1 \))

### Step 2: Rearrange the formula to solve for surface tension \( T \)
Rearranging the formula gives:

\[ T = \frac{h \cdot \rho \cdot g \cdot r}{2 \cos \theta} \]

### Step 3: Substitute the known values into the equation
Substituting the known values into the equation:

\[ T = \frac{0.15 \, \text{m} \cdot 900 \, \text{kg/m}^3 \cdot 10 \, \text{m/s}^2 \cdot 0.00015 \, \text{m}}{2 \cdot 1} \]

### Step 4: Calculate the surface tension \( T \)
Calculating the numerator:

\[ 0.15 \cdot 900 \cdot 10 \cdot 0.00015 = 0.002025 \, \text{N/m} \]

Now, divide by 2:

\[ T = \frac{0.002025}{2} = 0.0010125 \, \text{N/m} \]

### Step 5: Convert to milliNewtons per meter
To convert to milliNewtons per meter:

\[ T = 0.0010125 \, \text{N/m} \times 1000 \, \text{mN/N} = 1.0125 \, \text{mN/m} \]

### Step 6: Round to the nearest integer
Rounding \( 1.0125 \) to the nearest integer gives:

\[ T \approx 1 \, \text{mN/m} \]

### Final Answer
Thus, the surface tension of the liquid is approximately **1 mN/m**.